There is the nice integral by V. Reshetnikov, $$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\;\alpha}\tag1$$
also discussed in this post. By direct analogy, we can consider its cousin,
$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\,\beta^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2}\;\beta}\tag{2a}$$
and the version investigated also by Reshetnikov,
$$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}\tag{2b}$$
Given some integer/rational $N$, it entails finding algebraic numbers $\alpha, \beta, \gamma$ such that,
$$\begin{aligned} \frac{1}{N}=I\left(\alpha^2;\ \tfrac12,\tfrac13\right)= \frac{B\left(\alpha^2;\ \tfrac12,\tfrac13\right)}{B\left(\tfrac12,\tfrac13\right)} \end{aligned} $$
$$\begin{aligned} \frac{1}{N}=I\left(\beta^2;\ \tfrac12,\tfrac14\right)= \frac{B\left(\beta^2;\ \tfrac12,\tfrac14\right)}{B\left(\tfrac12,\tfrac14\right)}\\ \end{aligned}$$
$$\begin{aligned} \frac{1}{N}=I\left(\gamma^2;\ \tfrac14,\tfrac14\right)= \frac{B\left(\gamma^2;\ \tfrac14,\tfrac14\right)}{B\left(\tfrac14,\tfrac14\right)}\\ \end{aligned}$$
with regularized beta $I(z;a,b)$, incomplete beta $B(z;a,b)$, and beta function $B(a,b)$. Solutions for $\alpha, \beta, \gamma$ for $N=2,3,5,7,11$ are known, for the latter two as,
$$\begin{array}{|c|c|c|} \hline N & P(\beta)=0 &P(\gamma)=0 \\ \hline 2 & -4 + 4 \beta^2 + \beta^4 & -1 + 2 \gamma^2\\ 3 & -2 - 2 \beta + \beta^2 & - 1 + 2 \gamma + 2 \gamma^2\\ 5 & -4 + 8 \beta + 4 \beta^2 - 8 \beta^3 + \beta^4 & - 1 + 8 \gamma - 4 \gamma^2 - 8 \gamma^3 + 4 \gamma^4\\ \hline \end{array}$$
with $N=7$ a $12$-deg, and $N=11$ a $30$-deg, both reciprocals. (Polynomials for $\alpha$ are given here.)
Q: For $\beta$ and $\gamma$, why are they roots of reciprocal polynomials when $N=3,5,7,11$ (and presumably others), but different polynomials when $N=2$?
Ex. For $N=5$ and using $(2a)$ yields the succinct $\beta = 1-\tan\tfrac{3\pi}{20}$ hence, $$\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt[4]{1-x}\ \sqrt{1-x\big(1-\tan\tfrac{3\pi}{20}\big)^2}}=\frac{\sqrt{2}\,\pi}{5\big(1-\tan\tfrac{3\pi}{20}\big)}$$
